ATM conference 2015

The Inquiry Maths session at the Association of Teachers of Mathematics conference 2015 attracted 40 participants. We looked at the intersecting sequences prompt, coming up with a wide variety of questions and observations:

What other sequences have 7 and 13 in? Is it a coincidence that 7 and 13 are prime?

The sequence generated by n + 6 also contains 7 and 13.

Another sequence with 7 and 13 is generated by 1.5n - 2.

Are there other numbers that are in both sequences?

How does the red sequence lead to the red expression?

What is the link between the two expressions and the last one?

The product of the coefficients of n in 3n - 2 and 2n + 1 give the 6, but I don’t know how to find the constant.

The differences between each term in the first two sequences are -2 -1 0 1 2 and the expression for the nthterm of that sequence is n - 3.

If you sum the three expressions 3n - 2, 2n + 1, and n - 3, you get 6n - 4. The coefficient is correct, but not the constant.

Looking at the numbers between 7 and 13, there are none in the red sequence, one in the blue, and two in the green. What would the expression be for the sequence with three numbers between 7 and 13?

If you arrange the expressions for the nth terms vertically, what expressions would come before and after the ones in the prompt?

Josh Evans (a teacher from Steyning Grammar School, UK) showed us how the lowest common multiple could be used to find the coefficient of n. 3n - 2 and 5n + 3 lead to 15n - 2 and the lcm(3,5) = 15. Similarly 2n + 1 and 4n + 3 give 4n + 1 and the lcm(2,4) = 4. He speculated that the constant comes from one of the first two expressions (-2 and +1 in his examples). A counter-example was produced by another participant.

The second contributor showed how any expression with a coefficient for n of 1, 2, 3, or 6 would give a sequence including 7 and 13. This can be achieved by ‘sliding’ the multiplication table until it gives 7 and 13. For example, 2n gives 2 4 6 8 10 12 14. You could slide this 5 to the right (2n + 5) or 3 to the left (2n - 3) and so on.

Finally, Anna Dickson (a teacher who now works for the Oxford University Press) showed how she had started to use modular arithmetic to generate expressions for sequences that start with 7. She demonstrated her work using modular “clocks”:

(6/1)n + 1 where 1 = 1 (mod 6)

(6/2)n + 4 where 4 = 1 (mod 3)

(6/3)n + 5 where 5 = 1 (mod 2)

(6/4)n + 5.5 where 5.5 = 1 (mod 1.5)

This generalises to mn + 1 (mod m) where 7 - m = 1 (mod m). Therefore, we have a general expression to create a sequence that starts with 7: mn + (7 - m).

The ATM conference 2015 (March 30 - April 2) was held in Daventry, UK. Its theme was "Thinking Mathematically".The picture shows participants in the Inquiry Maths session.